The light of two different frequencies whose photons have energies 3 . 8 eV and 1 . 4 eV respectively, illuminate a metallic surface whose work function is 0 . 6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

Step-by-step solution

Expression for maximum kinetic energy can be written as, K E m a x = E - ϕ K E m a x ,   1 = E 1 - ϕ = 3 . 8 - 0 . 6 = 3 . 2   eV , K E m a x ,   2 = E 2 - ϕ = 1 . 4 - 0 . 6 = 0 . 8   eV K E m a x ,   1 K E m a x ,   2 = 3 . 2 0 . 8 = 4 ⇒ 4 = 1 2   m v 1 2 1 2   m v 2 2 ⇒ v 1 2 v 2 2 = 4 ⇒ v 1 v 2 = 2 : 1